If α ,β ,gamma are roots of x^3 - x - 2 = 0 , then value of α^5 + β^5 + gamma^5 is-

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4-2.Quadratic Equations and Inequations

normal

If $\alpha ,\beta ,\gamma$ are the roots of $x^3 - x - 2 = 0$, then the value of $\alpha^5 + \beta^5 + \gamma^5$ is-

$5$

$8$

$9$

$10$

Solution

$\alpha \,is\,a\,root\, \Rightarrow \,{\alpha ^3} = \,\alpha + 2 $
$ \Rightarrow {\alpha ^5} = {\alpha ^3} + 2{\alpha ^2} = \alpha + 2 + 2{\alpha ^2} $
$ \therefore {\alpha ^5} + {\beta ^5} + {\gamma ^5} = 2({\alpha ^2} + {\beta ^2} + {\gamma ^2}) + ({\alpha ^{}} + \beta + \gamma ) + 6 $
$ = 2(\sum \alpha – 2\sum \alpha \beta ) + \sum \alpha + 6$
$ = 2(0 – 2( – 1)) + 0 + 6 = 10 $

Standard 11

Mathematics

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normal

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hard

(JEE MAIN-2021)

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If $\alpha ,\beta ,\gamma$ are the roots of $x^3 - x - 2 = 0$, then the value of $\alpha^5 + \beta^5 + \gamma^5$ is-

Solution

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